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docker41 [41]
2 years ago
6

957.75 + 83.86 the sum is

Mathematics
2 answers:
pishuonlain [190]2 years ago
8 0

Answer:

1,041.61

Simply stack the numbers (lining up the decimals) and add right to left.

FrozenT [24]2 years ago
4 0
The sum for this answer would be 1,041.61
You might be interested in
Find the perimeter of a rectangle with a length of 6x+3 and a width of -2x-5.
blsea [12.9K]
The simple way of calculating a perimeter is to add all of the sides.
For a rectangle this would equal 2 x length + 2 x width
P = 2(6x + 3) + 2(-2x - 5)
= 12x + 6 - 4x - 10
= 8x - 4

At a higher level both the length and width must be greater than zero (= zero is a trivial rectangle)
6x + 3 > 0
6x > -3
x > -0.5

-2x - 5 > 0
2x + 5 < 0 (multiplying by -1 reverses the inequality)
2x < -5
x < -2.5

This rectangle cannot exist as x cannot be < -2.5 and > -0.5 at the same time!
8 0
3 years ago
Read 2 more answers
What's the distance between (-2, 5) and (8, 5)
Wewaii [24]

Answer:

10

Explanation:

The distance between two points of coordinates (x1, y1) and (x2, y2) can be calculated as:

\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}

So, the distance between (-2, 5) and (8, 5) is equal to:

\begin{gathered} \sqrt[]{(8-(-2))^2+(5-5)^2} \\ \sqrt[]{(8+2)^2+(0_{})^2} \\ \sqrt[]{(10)^2+0} \\ \sqrt[]{100} \\ 10 \end{gathered}

Then, the distance between (-2, 5) and (8, 5) is 10.

7 0
2 years ago
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
How do I do this ffs
lidiya [134]

Answer:

I am not sure but for me the answer is 1

6 0
3 years ago
Can you please explain Number 7 with steps
Marizza181 [45]

Answer:

this is not exact because i Estimated the square root of 33

Step-by-step explanation:

4 0
4 years ago
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